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Driveby
Posts:
167
Registered:
2/9/07
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Re: stuck proving trig identity
Posted:
Jun 4, 2007 5:33 PM
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On Mon, 04 Jun 2007 20:39:53 -0000, Russell <russell@mdli.com> wrote:
>On Jun 4, 10:04 am, "Stephen J. Herschkorn" <sjhersc...@netscape.net>
>wrote:
>> conrad wrote:
>> >(cot x - 1)/(cot x + 1) = (cos 2x) / (1 + sin 2x)
>> >=> (cos x - sin x) / (cos x + sin x) = (cos^2 x - sin^2 x) / (1 + 2sin
>> >x cos x)
>>
>> >This is where it gets real messy. So I must be missing
>> >something obvious at this point. Any hints?
>>
>> Conrad has been given an awful lot of advice that is way too complicated
>> here. The first response was right on the mark:
>>
>> Paul Sperry wrote:
>> >cos(2x) = .....;
>> >sin(2x) = .....
>
>But didn't the OP already write out correct formulas
>for those in his second line? Maybe you're thinking of
>different identities than I am; in any case I see no
>particular reason for the OP to abandon his original
>approach. He should now focus on the denominator
>of his 2nd-line RHS, and rewrite the 1 using the world's
>best-known trig identity. Then, factorize top and bottom.
>Not messy at all!
This is the identity the OP wants to prove
(cot x - 1)/(cot x + 1) = (cos 2x) / (1 + sin 2x)
He got the LHS to this point
(cos^2 x - sin^2 x) / (1 + 2 sin x cos x)
Now, substiture the obvious double-angle idetities into the LHS and
proof done.
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